Let $R$ be the region enclosed by the $x$ -axis, the line $x=1$, the line $x=3$, and the curve $y=\dfrac1x$. $y$ $x$ ${y=\dfrac 1x}$ $ R$ $ 1$ $ 3$ A solid is generated by rotating $R$ about the $x$ -axis. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Solution: Let's imagine the solid is made out of many thin slices. $y$ $x$ ${y=\dfrac 1x}$ Each slice is a cylinder. Let the thickness of each slice be $dx$ and let the radius of the base, as a function of $x$, be $r(x)$. Then, the volume of each slice is $\pi [r(x)]^2\,dx$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [r(x)]^2\,dx$ This is called the disc method. What we now need is to figure out the expression of $r(x)$ and the interval of integration. Let's consider one such slice. $y$ $x$ ${y=\dfrac 1x}$ $ 1$ $ 3$ $r$ The radius is equal to the distance between the curve $y=\dfrac1x$ and the $x$ -axis. In other words, for any $x$ -value, $r(x)=\dfrac1x}$. Now we can find an expression for the area of the cylinder's base: $\begin{aligned} &\phantom{=}\pi [r(x)}]^2 \\\\ &=\pi\left(\dfrac1x}\right)^2 \\\\ &=\dfrac{\pi}{x^2} \end{aligned}$ The leftmost endpoint of $R$ is at $x=1$ and the rightmost endpoint is at $x=3$. So the interval of integration is $[1,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_1^3 \left(\dfrac{\pi}{x^2}\right)dx \\\\ &=\pi \int^3_1 x^{-2}\, dx \end{aligned}$ Let's evaluate the integral. $\pi \int^3_1 x^{-2}\, dx=\dfrac{2\pi}{3}$ In conclusion, the volume of the solid is $\dfrac{2\pi}{3}$.